FRM Fun 4

In the casino game of craps, the player rolls two six-sided dice. If the first roll is a 7 or 11, the player wins; if the first roll is a 2 or 12, the player loses. Normally, of course, the dice are independent and the expected outcome (mean) of rolling both dice is a seven (7). However, a crooked casino inserts magnets into the dice! With the hidden magnets, although each individual die remains a uniform distribution (i.e., probability = 1/6), the two dice are no longer independent but now have a positive correlation of + 0.50. The casino does this to make a roll of 2 (double ones) or 12 (double sixes) more likely.

Questions:

Does the expected sum of the two dice change, and if so, to what?

Does the variance of the sum of the two dice change, and if so, to what?

E(X)=E(Y)=3.5
E(X+Y)=7; Var(X)=sigma([X-E(X)]^2)/(6-1)=3.5 similarly Var(Y)=3.5 and corr(X&Y)=0 =>Var(X+Y)=Var(X)+Var(Y)=3.5+3.5=7
When X and Y became dependent with corr.=.5 then it does not affect the expected value but only the relative outcomes of X and Y. Also since std. deviation of Expected values does not change and values remains same.That is when 2 occurs on dice 1 then 2 also occurs on die 2 similarly the case for six for six. Hence E(X+Y)=7 does not changes.
Cov(X,Y)=corr*sqrt(Var(X)*Var(Y))=.5*sqrt(3.5*3.5)=1.75
Var(X+Y)=Var(X)+Var(Y)+2*Cov(X,Y)=3.5+3.5+2*1.75=7+3.5=10.5so the total variance changes by 10.5-7=3.5

ShaktiRathore: your is partly correct, so we are entering you once also into the weekly drawing. The (n-1) is tempting because we would use (n-1) adjustment if we were extracting a sample variance from an actual trial (e.g., an observed simulation). But, in the case of a six-sided die, we know the "parametric" distribution ex ante, so it's not a sample: we don't need the (n-1). To put this another way, assume we did not know the underlying uniform distribution, and we only observed the following set of outcomes: {1, 2, 3, 4, 5, 6} ... in this case, the (n-1) is appropriate and we should compute a sample variance = [(1-3.5)^2+(2-3.5)^2+(3-3.5)^2+(4-3.5)^2+(5-3.5)^2+(6-3.5)^2]/(6-1) = 3.5 .... why? b/c we don't actually know the population mean of 3.5 we consume a d.f. using the sample mean of 3.5. But a single-six sided die is not a sample, we know the distribution that characterizes it.

This problem employ two common, useful FRM formulas:

Variance (X) = E[X^2] - [E(X)]^2. As shown above, it easily gives us the variance of six-sided die. Also, it can give us the variance of a Bernoulli (EDF or default).